# Continuous maps in topology; the definition?

I am just wondering, given the definition of continuous maps as follows,

A functionn $f:X \to Y$ is continuous if for every open subset $U$ of $Y$ the preimage $f^{-1}U$ is open in $X$.

I guess mathematically, this doesn't necessarily mean that "an open subset of $X$ is mapped to an open subset in $Y$"?

It's only that the open subset of $Y$ must originate from an open subset in $X$, but not necessarily that every open $V$ of $X$ will be mapped to some open $U$ of $Y$.

Is this understanding correct?

• That's right. A constant function is continuous, but for most topologies does not map an open set to an open set. For a familiar somewhat different example, the image of $(0,42)$ under the sine function is the non-open set $[-1,1]$. May 6, 2016 at 22:37
• It means that, if $u$ is an open subset of $Y,$ then $f^{-1}(U)$ is open in $X.$ I don't know if that means that $U$ "originates from an open set in $X.$" What does "originate from" mean???
– bof
May 6, 2016 at 22:42
• Oh, I meant to say that if $f(x) \in Y$ then $f(x)$ originates from the open subset $x$ is in $X$. The point of the question was, do all open sets in $X$ get mapped to open sets in $Y$ if $f$ is continuous? But I guess not, looking at the answers! May 6, 2016 at 22:45
• Thanks, but your reply explains nothing. What does "$f(x)$ originates from the open subset $x$ is in $X$" mean? I can't even parse that, and I don't what "originates" means mathematically.
– bof
May 6, 2016 at 22:51
• I might want to possibly change that wording then, I didn't really entirely have any rigid mathematical definition behind it, it was more of an intuitive thing to try to convey my question colloquially...do you at least have an intuition or feel of what I was trying to say there? If so, how better would you word it mathematically? If there's a suggestion, I'd change it May 6, 2016 at 23:24

Yes, that is correct.

A function that maps open sets to open sets is called an open map, i.e a function $f : X \rightarrow Y$ is open if for any open set $U$ in $X$, the image $f(U)$ is open in $Y$.

Open maps are not necessarily continuous.

Then there is the concept of closed maps which maps closed sets to closed sets. A map may be open, closed, both, or neither and continuity is independent of openness and closedness.

A continuous function may have one, both, or neither property.

• Right, so $f$ being continuous doesn't necessarily mean that open sets from $X$ goes to open sets in $Y$. And if it does, it's another thing called the open map...you say? Actually, then can I ask for a step further; if $f$ is a homeomorphism, then it necessarily maps open sets in $X$ to open sets in $Y$, yes? May 6, 2016 at 22:47
• @JohnTrail Yes. May 6, 2016 at 22:52
• @JohnTrail Yes to your second question too. If $f$ is a homeomorphism then the inverse $f^{-1}$ is continuous which means $f$ is an open map. May 6, 2016 at 22:59
• Thank you, made things clearer now :) May 6, 2016 at 23:23

Continuous maps don't have to map open sets to open sets. An example is the map $f:\mathbb{R}\rightarrow\mathbb{R}$ given by $f(x)=x^2$ which maps $(-1,1)$ to $[0,1)$ which is not open and not closed.

Yes it is. Consider, for example, the continuous function $$f(x) = x^2$$ What is the image of the open set $(-1,1)$ ?

Yes. I tend to think of this (quickly) as "continuous = open sets come from open sets" and more slowly as "the preimage of an open set is an open set".

There is another term: A map is open if it takes open sets to open sets. A map may be open, continuous, neither, or both. This is the other idea that a person first learning this definition of continuous may conflate with continuity.

To help set this idea. The sine map on $\Bbb{R}$ is clearly continuous (using the ideas you had before topology). However, the interval $(-100,100)$ (large enough easily to hold an entire period) is mapped by sine to $[-1,1]$, so continuous maps are not automatically open maps.

• I didn't understand the question. I look at the answers and everyone agrees with the OP's statement, without explaining what it means. What does it mean for one set to "originate from" or "come from" another? Suppose $f:\mathbb R\to\mathbb R^2$ is a constant function. Suppose $U$ is the open unit disk in $\mathbb R^2.$ What open set in $\mathbb R$ does $U$ "come from"?
– bof
May 6, 2016 at 22:48
• @bof : That would be $f^{-1}(U) = \{x \in \Bbb{R} \mid f(x) \in U\}$, the collection of points that $f$ takes to $U$. May 6, 2016 at 22:49
• I see. My problem was that, given $f:X\to Y$ and subsets $A\subseteq X,B\subseteq Y,$ I was unsure of the meaning of the strange expression "$B$ comes from $A$" or, in the OP's version, "$B$ originates from $A$". If I had to guess, I would have guessed that it means that $B=f(A).$ From your answer, I now know that it means $A=f^{-1}(B).$ Thanks.
– bof
May 6, 2016 at 23:15
• @bof. Yes. It is worth remembering that these maps need not be injective, so a single point may have multiple preimages. I.e. $f^{-1}(y)$ may be several points, or a whole set. Consider the function $f:\Bbb{R}^2 \rightarrow \Bbb{R}: x \mapsto 0$. The preimage of $\{0\}$ is $\Bbb{R}^2$. The preimage of other points of $\Bbb{R}$ is the empty set. We're working with set-valued functions here. May 6, 2016 at 23:31

Yes that is correct.

If a function maps open sets to open sets, then it is said to be an open map.

A continuous map is not necessarily open. For example the $\sin$ function is continuous but not open since it maps the open interval $(0,\pi)$ to $(0,1]$, which is not open.

However, note that if a continuous map $f$ has an inverse $f^{-1}$, then $f^{-1}$ is an open map.